Could you give this a try in my (smichr) improve_relational branch? I saw
that error there, too, and a correction I made avoided the error.
/c
On Tuesday, December 2, 2014 8:44:33 AM UTC-6, bruno....@gmail.com wrote:
>
>
> Hi all,
>
> I am new to sympy (and to symbolic computations...) and I meet a problem
> for solving (possibly difficult) inequalities depending upon on real
> symbol
>
> (I have used k = symbols('k',real=True,positive=True)).
>
> First I build a set of (5) inequalities, stored in a list L and I want
> to find
> for which subset (of real numbers) they are all true. Substition for a
> given value
> of the symbol k works, for instance :
>
> for elem in L:
>
> print(elem.subs(k,Rational(9,4)))
>
>
> outputs five True. But solving this way :
>
>
> res = solve([k>2, k <= 3, L[0], L[1], L[2], L[3], L[4]], k)
>
>
> leads to the following error:
>
> Traceback (most recent call last):
>
> File "<stdin>", line 1, in <module>
>
> File "/usr/lib/python2.7/site-packages/sympy/solvers/solvers.py", line
> 691, in solve
>
> return reduce_inequalities(f, symbols=symbols)
>
> File "/usr/lib/python2.7/site-packages/sympy/solvers/inequalities.py",
> line 513, in reduce_inequalities
>
> poly_reduced.append(reduce_rational_inequalities([exprs], gen))
>
> File "/usr/lib/python2.7/site-packages/sympy/solvers/inequalities.py",
> line 255, in reduce_rational_inequalities
>
> result = solution.as_relational(gen)
>
> File "/usr/lib/python2.7/site-packages/sympy/sets/sets.py", line 1212, in
> as_relational
>
> return Or(*[set.as_relational(symbol) for set in self.args])
>
> AttributeError: 'Complement' object has no attribute 'as_relational'
>
>
>
> From these messages, at first glance, I guessed for something wrong with
> my relationals L but
>
> each element of L seems to have a good type:
>
> >>> type(L[1])
>
> <class 'sympy.core.relational.StrictGreaterThan'>
>
>
> the same than what returns: type(k>2). So I am wondering if I have made
>
> some mistake or if this is due to some limitations as my relations are
> quite complicated :
>
> >>> L[1]
>
> 4*(-k**6 + 2*k**4 + 4*k**2 + 12)/(k*(-k**8 + 2*k**6 + 4*k**4 + 16*k**2 -
> 8)) > (k**8 - 4*k**6 + 16)/(k*(k**8 - 2*k**6 - 4*k**4 - 16*k**2 + 8))
>
>
> Thanks for any help or advice
> Bruno
>
>
>
>
>
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